Topic open: Electric fields, potentials and capacitors

The following topic is now open for discussion: Electric fields, potentials, capacitors and parallel plates

If you have a question regarding this topic, please post a comment to this post by clicking on the comment link below.

Topic open: Circuits

The following topic is now open for questions: Circuits. Anything about resistors, capacitors, RC, RL, RLC circuits.

To pose a question, please post a comment to this post by clicking on the comments link below.

Question: Two resistors, Two capacitors (Ch 19 #50)

Ch 19 #50 part b

I am not sure how to distinguish whether a capacitor is charging or discharging from this problem. Sometimes, I can't tell through which resistor will a capacitor will discharge...is it through the one in parallel or in series?

Let's just look at simplifying what we have when the switch is open, then closed.

When the switch is open there are two resistors in series which are in parallel with two capacitors in series. The voltage across the set of resistors will be the same as the voltage across the set of capacitors (24 V in this case). At point "a" there will be some voltage/potential between 0 and 24 V due to the voltage dropped (IR) across one of the two resistors. Likewise at point "b" there will be some voltage/potential between 0 and 24 V as there is some voltage dropped (Q/C) across one of the two capacitors. (Note that to determine this you will have to use the equivalent resistances and capacitances of each branch of the circuit to determine the net current and charge, respectively.)

Now, when the switch is closed, charge will flow from a to b or vice versa because the potential at a and b were not equal initially. Once the switch is closed this potential difference cannot be maintained. Without working through to see which is greater and thus which direction the charge will flow I can't say whether the capacitors are charging or discharging. However, I can tell you if the potential at "a" is greater than at "b" the current (charge) will flow toward the capacitors and thus the capacitors will charge (partially). Conversely if the potential at "b" is greater than at "a" the current (charge) will flow away from the capacitors and the capacitors will discharge (partially).

Good luck. This is a tricky problem.

Question: pulling apart capacitor plates

A parallel-plate capacitor with plate area 2.0 cm^2, air-gap .5mm, is connected to a 12V battery, if the air-gap is pulled to .75mm, how much work is required?

calculated Q=4.2E-11 C, V(d=0.75)=18V, delta PE=Q*delta V = 2.52E-10 J, but the answer is 1.3E-10 J, which could be a half of my answer. PE/2? why? Would you please explain this to me? Thanks

Ok, so the thing you are changing when you pull a parallel plate capacitor apart is the capacitance (it's ability to store energy).

The capacitance depends on distance like:

C=εo*(A/d)

so clearly the capacitance will decrease if the distance between the plates is increased. Let's consider the capacitance before (C), and after (C'):

C=εo*(A/d)
C'=εo*(A/d')

Ok. So now let's consider how this will effect the energy stored in the capacitor. This can be written in 3 different ways (by substituting Q=CV into the first):

PE=(1/2)QV=(1/2)CV²=(1/2)Q²/C

Which of these should we use? Well, since the plates are connected to a 12 V battery, the voltage will be the same before and after moving the plates. Note that the charge may change so we can't use the expressions with Q in them because we don't know how Q is changing (ok, yes, we could figure it out, but it would be equivalent to using the second expression, so let's not bother).

All right, so let's put it together:

W=-ΔPE=-((1/2)C'V²-(1/2)CV²)
W=-((1/2)εo*(A/d')V²-(1/2)εo*(A/d)V²)
W=-εo*A*V²/2((1/d')-(1/d))

Plugging in the numbers given in the equation above for the work done, I get: 8.49e-11J (using 12V). You also write 18V in your question, with that number I get an answer: 1.91e-10J. Sorry to muddy the waters with more answers... the method should be correct. (maybe you can recheck which numbers the question uses?)

If you haven't already, I'd suggest taking a look at the similar example (though with constant charge not voltage) posted on the main page: Ch. 17, #50.

Hope that helps. Feel free to ask questions by posting comments.

Question: difference between RC, RL, LC

How do you tell the difference between RC, RL, LC circuits?

Each of these circuits, and what they qualitatively do in DC and AC circuits...

RC - a resistor and capacitor in series. Exhibits charging behaviour with characterisitic time constant with DC voltage source. Acts as a high pass filter (allows high frequency currents, but not low frequency currents) in AC circuits.
RL - a resistor and inductor in series. Acts as a short with a DC voltage source, but smooths out rapid variations in current. Acts as a low pass filter (allows low frequency currents, but not high frequency currents) in AC circuits.
LC (and RLC) - an inductor and capacitor (and resistor) in series. If initially charged, has oscillitory behaviour (damped if also has a resistor). Has resonant behaviour with AC driving voltage (damped if also has a resistor).

Question: LC in parallel

How to calculate the resistance of a capacitor and an inductor connected in parallel?
(-XcXL/(XL-Xc))^2 ?

This is a little bit trickier that considering elements in series because the inverse of the impedances have to be added. I'm pretty sure it is beyond the scope of this course to need to do this calculation, but I will put it here anyway. I'm also not sure how you would arrive at the answer without using the appropriate complex represenations of the impedance, so I will use them but try to be careful to explain.

So, for elements in parallel:

1/Z_tot=1/Z_C+ 1/Z_L+ 1/Z_R ...[1]

We will consider just an LC circuit (with ideal components), so R=0. Now, in complex notation Z_Cand Z_L are:

Z_C=1/iωC ...[2]
Z_L=iωL ...[3]

So substituting [2,3] into [1]:

1/Z_tot=iωC+1/iωL

or, using the fact that 1/i=-i:

1/Z_tot=iωC-i/ωL
1/Z_tot=i(ωC-1/ωL) ...[4]

so, we can now replace 1/ωC with X_C and ωL with X_L:

1/Z_tot=i(1/X_C-1/X_L) ...[5]

Now we can simplify the part in the brackets, giving us:

1/Z_tot=i(X_L-X_C)/X_CX_L ...[6]

or taking the inverse of the fraction on both sides (and once again using 1/i=-i):

Z_tot=-iX_CX_L/(X_L-X_C) ...[7]

Now, if we had also had a resistor in parallel in the circuit (R), we would end up at equation [7] with both real and imaginary parts. To find X_tot we would have to find the magnitude of the resultant phasor. Since Z_tot is purely imaginary, we need only to find the magnitude of this number, which we can express as:

X_tot=-X_CX_L/(X_L-X_C) ...[8]

Which is pretty much what I think you suggested. Note that if we hadn't used the imaginary numbers we would not have been able to get the right signs and would have come up with a different answer.

Looking back at eqn. [7], let's think about what this represents... In the imaginary plane Z_tot would be a phasor pointing along the y-axis. For values of X_L and X_C which make Z_tot positive, the impedance could be considered "inductive", and inversely if Z_tot were negative, the impedance could be considered "capacitive".

Let's consider two extreme cases...

If the frequency is high, X_L is very large, and X_C is very small (let's say negligible). In the numerator of Z_tot the frequency cancels out, so this is a constant. If we look at the denomenator, X_L >> X_C, so the denomenator is approximately X_L. Since X_L is large, Z_tot will be small and negative (capacitive). In fact if you substitute in equations [2,3] ignoring X_C in the denominator, you will recover Z_C. This is consistant with the inductor providing a break in the circuit, and the capacitor providing a short.

Similarly, if the frequency is very low, X_C becomes very large, and X_L very small. In this case, the result is small and positive (inductive), and indeed again, if you work it out you will recover Z_L as Z_tot. This is consistant with the capacitor providing a break in the circuit, and the inductor providing a short.

To summarize:

ω→∞ Z_tot→Z_C and current flows through the capacitor
ω→0 Z_tot→Z_L and current flows through the inductor

These limiting cases can be worked out simply as a thought experiment (consider which branch of the circuit has a small resistance and which has a large resistance) and are often more insightful than getting through the algebra.

Hope this helps.

Topic Open: RC, RL, LC, LRC circuits

The following topic is now open for questions: RC, RL, LC, LRC circuits (and any AC circuits)

To pose a question, please post a comment to this post by clicking on the comments link below.

Question: Increase plate separation, increase energy stored in capacitor

If two plates of a capacitor with constant charge have their separation doubled, the energy stored also doubles. I'm confused as to why this happens, since I thought E-stored was related to C, and C decreased as distance of separation increased. I'm looking at the equation Energy Stored = 0.5CV^2.

You are right that the energy stored in a capacitor if related to C, the capacitance. However, you have to consider what is being held constant as the separation doubles. If you look at the equation: U=(1/2)CV², you will notice that U is proportional to C, however you have a factor of V² that changes in a way you can't determine directly. If instead, you re-write the equation substituting in V=Q/C, so that you have: U=(1/2)Q²/C, then you know that Q is a constant, so we can talk about U being proportional to 1/C. Thus, as C decreases, U increases.

It might help you to look at the example I posted on this if you haven't already: Chapter 17, #50. Note also that since the amount of energy stored in the capacitor is increased, work must be done to increase the separation between the plates.

Question: Capacitor miscellany

here are my questions about capacitor:

1 #51 b)Why doubling?

The energy stored in a capacitor is given by:

U=(1/2)QV=(1/2)CV²=(1/2)Q²/C.

In Ch. 17, #51 (b) we are told that the amount of charge on each plate is doubled (ie. Q is doubled), and the capacitor remains connected to a battery (ie. V is held constant). So, we can look at an initial energy, U₀, and a new energy after the charge is doubled, U':

U₀=Q₀V
U'=Q'V=(2Q₀)V

The ratio of the new energy to the old is then:

U'/U₀=Q'V/Q₀V=2

So, indeed if the charge on the capacitor is doubled, the energy stored in the capacitor is doubled.

2 Say, we have 2+Q on the left plate and 2 -Q on the rite, can I just add 1 +Q to the left and add nothing to the rite? Or it is a must for me to add the same amount of charges to both sides?

You must add the same amount of charge to both sides. (Check out the Ch. 17 summary on where it describes a capacitor.) If you added more of one charge to one side than the other, the net charge of the capacitor would be non-zero. I'm not sure what exactly would happen, but it wouldn't be a simple capacitor anymore and wouldn't follow the behaviour we've been discussing. Sorry I can't give a better explanation than that.

3 If I added 1 +Q to left and 1 -Q to rite, the total charge should be +2Q, rite?

Well, the net charge would be zero, the total amount of charge would be |2Q| (those || are to show we are talking magnitude/absolute value and disregarding the signs), but the charge on the capacitor (the "Q" you use in the capacitor equation) would be 1Q. The measure of charge that matters in a capacitor is how much you have separated off to each plate, so it is this charge you consider.

4 #77b) what i did was PE/2=1/2*m*v^2, what is wrong with my solution( i know it must be wrong -_-||)?

I had a student ask this question last year, I think the answer given there will help you also. Check out electrostatics: previous student questions, it is the second item on the page.

I believe your problem lies in the factor of 1/2 you are using for the potential energy... I'm not sure where this comes from, and I'm pretty sure it isn't necessary. ;)

Good Luck, hope this helps. As always you can ask follow-up questions.

Topic open: Anything Capacitor

The following topic is now open for questions: Capacitors.

To pose a question, please post a comment to this post by clicking on the comments link below.

Question: Back-to-Back Capacitors (Ch 17, #41)

One student asks: "I was able to find the number of charges using Q=CV...but I have no clue how to solve the question."

Another asks: "I figured out the total charge after charging to batteries, but I dunt understand why they made "the positive plates connected to each and the negative plates connected to each other" and I cannot get the answer either. Another question from #41: they said "the potential difference across reach". Does it have the same meaning of the voltage of a battery?"

The question states: "A 2.50-μF capacitor is charged to 857V and a 6.80-μF capacitor is charged to 652V. These capacitors are then disconnected from their batteries. Next the positive plates are connected to each other and the negative plates are connected to each other. What will be the potenial difference across each and the charge on each? [Hint: charge is conserved]"

Ok, so it seems there is some degree of confusion about this question and how to handle it. The first step is to calculate the charge on each isolated capacitor using: Q=CV, as suggested by the first student. The total charge when they are put together will be the sum of these two charges. You will have:

Q₁=(2.5e-6)(875)
Q₂=(6.8e-6)(652)
Q_total=Q₁+Q₂

The second question here actually holds a key part of the answer, and again, it is about interpreting the wording of a question. What does it mean when they say "the postive plates are connected to each other and the negative plates are connected to each other"? Let's draw this situation...

Notice that by connecting the terminals of the capacitors with like charge we have essentially placed them in parallel. So, we can treat them like parallel capacitors and find the equivalent capacitance of this arrangement. For capacitors in parallel, we have: C_eq=C₁+C₂.

Now we know the total charge on the two capacitors, and the equivalent capacitance of the two. We can now treat them as one capacitor with the total charge Q_total and equivalent capacitance C_eq to find the new voltage across both (note that the voltage across things in parallel is always equal).

Q_total=C_eqV_new
V_new=Q_total/C_eq

To address the second student's last question regarding the potential difference, the potential difference here is somewhat similar to the voltage across a battery. A battery's job is to raise the potential energy of charges passing through it, it puts more positive charge on one side and more negative charge on the other. The same is happening here, on one side of the two capacitors there is more negative charge and on the other there is more positive charge. So, we say there is a potential difference or a voltage across the two capacitors.

Hope this is helpful, if you have any follow-up questions please post a comment.